Question

help with creating equations for a horizontal ellipse and a horizontal hyperbola?

Answer 1

There are two fruit trees located at (3,0) and (–3, 0) in the backyard plan. Maurice wants to use these two fruit trees as the focal points for an elliptical flowerbed. Johanna wants to use these two fruit trees as the focal points for some hyperbolic flowerbeds. Create the location of two vertices on the y-axis. Show your work creating the equations for both the horizontal ellipse and the horizontal hyperbola. Include the graph of both equations and the focal points on the same coordinate plane.

Answer 2

I honestly did not get this lesson at all. Please explain how i can do this? Just get me to the point of having the equations written out and i should be ok. Please help :/ @jim_thompson5910 @phi

Answer 3

They give you the focii, but they don't give you a point on the ellipse or hyperbola

Answer 4

that's all i have. can it not be solved without?

Answer 5

you'd have to know at least one point on the ellipse to find the equation of it

Answer 6

It looks like you can choose any a, b so that it works

Answer 7

ok. so i can use any point?

Answer 8

focus is at (3,0) and (–3, 0)
the center is (0,0)

Answer 9

there are infinitely many ellipses that satisfy those 3 conditions

Answer 10

if you made something like (5,0) a point on the ellipse, then it would uniquely determine it

Answer 11

c is the distance from the center to one of the focus. c = 3

we use
c^2 = a^2 + b^2
or
9 = a^2- b^2
where a and b are the vertex distances from the center

Answer 12

so could a or b be anything that would fit in the equation acurately

Answer 13

I think it's c^2 = a^2 - b^2

Answer 14

yes. I would pick a=5 and b=4

Answer 15

'a' is the length of the semi-major axis

Answer 16

ok. 5 and 4 would make it easy enough.

Answer 17

What about the hyperbola?

Answer 18

see http://www.purplemath.com/modules/hyperbola.htm

now we want c^2 = a^2 + b^2
c=3, so we want
9 = a^2 + b^2

Answer 19

these variables represent the same things?

Answer 20

see the link. a is the distance from the center to the vertex

Answer 21

oh ok. sorry

Answer 22

and they go into the equation
x^2/a^2 - y^2/b^2 = 1

Answer 23

ok. what should i do with the variables?

Answer 24

Here is a graph of both

Answer 25

you pick the location of the vertex (it will be between 0,0 and 3,0)
i.e. the x value of the vertex will be between 0 and 3. In the plot, I picked 2
that makes a=2 and a^2 = 4
then use a^2 + b^2 = c^2
where c is 3 and a is 2 to find b, and then b^2

the equation will then be
x^2/a^2 - y^2/b^2 = 1

Answer 26

ok. i'm happy with those choices. Thank you so much for bringing me to these equations.